Question

Last two problems on the homework! please help with these 2. it helps when detailed explanations are given! thanks

A 2 kg ball is thrown horizontally, at 5m/s, from a height of 5 m. How fast will it be moving when it contacts the ground?

A 70 kg pole-vaulter converts the kinetic energy, of running at ground level, into potential energy to clear the crossbar at a height of 4 m above the gound. What is the minimum velocity that he must have had at takeoff (while on the ground) to clear the bar (minimum would be when velocity at the top was zero – you could still be moving forward at the top, but this would require more velocity at the beginning)?

Answer 1

ball's velocity 25 m/s when it hits the ground.
minimum velocity required is 8.85 m/s

Answer 2

how did you get those?

Answer 3

u = 5 m/s forever

v = -g t
h = 5 - 4.9 t^2

h = 0 at ground
so
4.9 t^2 = 5
t = 1.01 seconds

so v = - 9.81 * 1.01 = - 9.91 m/s

speed = sqrt (5^2 + 9.91^2) = 11.1 m/s
in other words I dissagree with jzee11

Answer 4

(1/2) m v^2 = m g h
v^2 = 2 gh
v = sqrt (2 g h)
= sqrt(2*9.81*4)
= 8.86 m/s

Answer 5

thank you so much!

Answer 6

You are welcome.